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In the renormalization procedure of quantum field theories, say $\lambda \phi^4$ theory for simplicity, we use the physical mass $m$, the physical coupling constant $\lambda$ and the physical field $\phi$ to fix the bare quantities $m_0$, $\lambda_0$ and $\phi_0$.

It is clear that the physical mass $m$ and the physical coupling constant $\lambda$ can be found from experiments (and using the S-matrix). But how does one "measure" the physical field $\phi$?

Edit: I would like to (hopefully) clarify my question. My question is related to field renormalization $\phi_0(x)=\sqrt{Z_\phi}\phi(x)$, and how we decide what value $\phi$ should be. For the physical mass and coupling constant, we can perform experimental measurements to determine their fixed values but we don't seem to do this for the physical field (as far as I understand; I still need to properly interpret JeffDror's answer). Are we basically looking at the divergence of Feynman diagrams and choose $Z_\phi$ such that we can absorb this divergence? If this is true, then this would mean that the "physical" field is a badly chosen word and it would make more sense to call it the "renormalized" field?

Hunter
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  • Do you mean the expectation value of $\phi$ in vacuum, $\langle 0|\phi|0\rangle$? – Isidore Seville Feb 25 '14 at 14:34
  • @IsidoreSeville I'm not sure to be honest... Is the VEV the way we renormalize the bare field $\phi_0$? – Hunter Feb 25 '14 at 14:37
  • won't field renormalization be related to the residue of a propagator? – innisfree Feb 25 '14 at 14:41
  • I think the answer depends on the context. If the said theory is used to model some condensed matter system, then it is possible to directly measure the v.e.v. – Isidore Seville Feb 25 '14 at 14:44
  • @innisfree as I understand it: we know from the LSZ formula in momentum space that the Green's function must a pole of the form $1/(p_i^2-m^2)$ for each external particle. This can be used to fix the physical mass. But I don't see how this can be used to determine the physical field. – Hunter Feb 25 '14 at 14:45
  • @IsidoreSeville thanks for the reply. I am hoping for an answer more related to quantum field theory in relation to particle physics. – Hunter Feb 25 '14 at 14:46
  • @hunter, there will be two relevant counter terms to fix, one with the pole of the propagator, the second with its residue. but this formalism doesn't result in anything measurable, afaik. – innisfree Feb 25 '14 at 14:51
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    I don't think @Hunter is referring to the VEV. The wavefunction renormalization arises from $\sqrt{ Z } = \left\langle k \right| \phi ( 0 ) \left| 0 \right\rangle $. This is equivalent to the residue of the propagator condition. – JeffDror Feb 25 '14 at 14:52
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    @Jeff can it be measured? only by turning off the vacuum? it can be calculated, if we know the cutoff, but i don't think that counts. if there is no cutoff it is infinite anyway so certainly not physical, i'd guess – innisfree Feb 25 '14 at 15:00
  • possible duplicate of In what sense is a scalar field observable in QFT? – joshphysics Feb 25 '14 at 17:46
  • @joshphysics thanks, your (nice) question is certainly related to mine, but I don't think it is a duplicate. My question is related to field renormalization $\phi_0(x) = \sqrt{Z_\phi} \phi(x)$, and how we decide what value $\phi$ should be. For the physical mass and coupling constant, we can perform experimental measurements to determine their fixed values but we don't seem to do this for the physical field (as far as I understand). – Hunter Feb 25 '14 at 18:10
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    @Hunter Fair enough! – joshphysics Feb 25 '14 at 18:16

1 Answers1

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To answer this we first need to be clear about why this wavefunction renormalization arises. For simplicity we focus on $\phi^4$ theory. For a free field we have, \begin{equation} \phi ( x ) \left| 0 \right\rangle = \int \frac{ d ^3 p }{ (2\pi)^3 } \frac{1}{ 2 E _{ {\mathbf{p}} } } e ^{ - i {\mathbf{p}} \cdot {\mathbf{x}} } \left| {\mathbf{p}} \right\rangle \end{equation} Apart from the factor of $ \frac{1}{ 2E _{ {\mathbf{p}} }} $ this is just equal to $ \left| {\mathbf{x}} \right\rangle $. Thus we can intereprate $ \phi ( {\mathbf{x}} ) $ acting on the vacuum as a field producing an $ {\mathbf{x}} $ eigenstate.

The equation above can be used to show that \begin{equation} \left\langle 0 \right| \phi ( {\mathbf{x}} ) \left| {\mathbf{p}} \right\rangle = e ^{ i {\mathbf{p}} \cdot {\mathbf{x}} } \end{equation} To continue to interpret $ \phi ( {\mathbf{x}} ) $ as the field that creates $ \left| {\mathbf{x}} \right\rangle $ eigenstates the renormalized field must obey this same relation. Thus we impose the renormalization condition: \begin{equation} \left\langle 0 \right| \phi _r ( 0 ) \left| {\mathbf{p}} \right\rangle = 1 \end{equation}
where $ \phi _r = \sqrt{ Z } \left( \phi - \left\langle \phi \right\rangle \right) $ is the renormalized field. If the wavefunction doesn't change under renormalization then $Z$ is just equal to $1$.

Now $ Z $ is not measurable for the same reason that the bare masses and couplings can't be measured. However, the renormalized field is measured in every process since if the condition above was not set to $1$ then we would be getting an extra factor for each external field line and another for each propagator in every diagram.

For example if we just forgot about the wavefunction renormalization we would get an extra $\sqrt{Z}$ for each external line and an extra $Z$ for each propagator.

JeffDror
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  • +1 Thanks for your reply. I'm going to have to think about it a bit more and maybe ask you some questions, before I can accept your answer. – Hunter Feb 25 '14 at 16:17